# Degrees of freedom of Sample Variance of Residuals (Chi-Square distribution?

In the context of jointly testing J linear restrictions, I am reviewing the distribution of the F-statistic, which is F(J, n-k). Below, R is a full row rank Jxk matrix, q is a Jx1 vector, and the null hypothesis tested is $$H_0:R\beta-q=0$$. We assume that residuals are homoskedastic and that they are conditionally normal.

I have that

$$F=(Rb-q)'[R(X'X)^{-1}R']^{-1}(Rb-q)/Js^2 \\=(Rb-q)'[R(X'X)^{-1}R']^{-1}(Rb-q)/J\sigma^2/(s^2/\sigma^2)$$ and this should be $$\sim{F(J, n-k)}$$

It should follow from the distributional result that, given two independent random scalars $$x_1\sim{\chi^2(p_1)}$$ and $$x_2\sim{\chi^2(p_2)}$$, it holds that

$$(x_1/p_1)/(x_2/p_2)\sim{F(p_1,p_2)}$$

In our case, we have that: $$x_1=(Rb-q)'{R(X'X)^{-1}R'(Rb-q)/\sigma^{2}}$$ $$p_1=J$$ $$x_2=s^2/\sigma^2$$ and then $$p_2$$ should be $$n-k$$. But here $$x_2$$ is not divided by $$n-k$$.

My intuition:

since we know that $$(n-k)(s^2/\sigma^2)\sim{\chi^2(n-k)}$$, it holds that $$s^2/\sigma^2\sim{\chi^2}(1)$$

In this way, $$p_2$$ would be just 1, when again it should be $$n-k$$. Can soebody shed some light on this?

• Oct 8, 2022 at 6:56

Let $$Q:= {(\mathbf R{\mathbf b }-\mathbf q)^\mathsf T\left[\mathbf R(\mathbf X^\mathsf T\mathbf X)^{-1}\mathbf R^\mathsf T\right]^{-1}(\mathbf R{\mathbf b }-\mathbf q)}.\tag 1$$

As outlined in this CV.SE post,

$$\frac Q{\sigma^2}\sim{\chi^2}^\prime\left[r(\mathbf R) ,\frac{{(\mathbf R{\mathbf b }-\mathbf q)^\mathsf T\left[\mathbf R(\mathbf X^\mathsf T\mathbf X)^{-1}\mathbf R^\mathsf T\right]^{-1}(\mathbf R{\mathbf b }-\mathbf q)}}{2\sigma^2}\right].\tag 2$$

Also,

\begin{align}\textrm{SSE}/\sigma^2 &\sim {\chi^2}_{n-r(\mathbf X) },\tag{3.1}\\(n-r(\mathbf X) )s^2/\sigma^2 &\sim {\chi^2}_{n-r(\mathbf X)}.\tag{3.2}\end{align}

Reformulating $$Q$$ and $$\textrm{SSE}$$ as quadratics in $$\mathbf y - \mathbf X \mathbf R^\mathsf T(\mathbf R\mathbf R^\mathsf T) ^{-1}\mathbf q,$$ it can be shown that they are distributed independently (cf.$$\rm [I],$$ section $$3.6,$$ pp. $$111-112$$).

Now \begin{align}F &:= \frac{Q/r(\mathbf R) }{\textrm{SSE}/[n-r(\mathbf X) ]}\\ &= \frac{(Q/\sigma^2)/r(\mathbf R) }{(\textrm{SSE}/\sigma^2)/[n-r(\mathbf X) ]}\\&\sim \mathsf F^\prime\left[r(\mathbf R),n-r(\mathbf X), \frac{{(\mathbf R{\mathbf b }-\mathbf q)^\mathsf T\left[\mathbf R(\mathbf X^\mathsf T\mathbf X)^{-1}\mathbf R^\mathsf T\right]^{-1}(\mathbf R{\mathbf b }-\mathbf q)}}{2\sigma^2}\right].\tag 4\end{align}

Therefore, under the null hypothesis $$\mathcal H_0: \mathbf R\boldsymbol\beta = \mathbf q,$$

\begin{align}F &= \frac{Q}{r(\mathbf R) s^2}\\ &\sim \mathsf F_{r(\mathbf R), n-r(\mathbf X) }.\tag 5\end{align}

## Reference:

$$\rm [I]$$ Linear Models, S. R. Searle, John Wiley & Sons., $$1971.$$